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The gestation time of humans has an approximate Normal distribution with a mean of 250 days and a standard deviatim- 6.0 days. A simple random sample of
n newborns is to be taken.
What is the minimum sample sized needed so that the sampling distribution of
bar(x) has a standard deviation of 0.5 day?
15
144
12
250

The gestation time of humans has an approximate Normal distribution with a mean of $250$ days and a standard deviatim- $6$ . $0$ days. A simple random sample of $n$ newborns is to be taken. $\newline$ What is the minimum sample sized needed so that the sampling distribution of $\bar{x}$ has a standard deviation of $0$ . $5$ day? $\newline$ $15$ $\newline$ $144$ $\newline$ $12$ $\newline$ $250$

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Find probabilities using the binomial distribution

Full solution

Q. The gestation time of humans has an approximate Normal distribution with a mean of

250

days and a standard deviatim-

6

.

0

days. A simple random sample of

n

newborns is to be taken.

\newline

What is the minimum sample sized needed so that the sampling distribution of

\bar{x}

has a standard deviation of

0

.

5

day?

\newline

15

\newline

144

\newline

12

\newline

250

Identify Formula: Step $1$ : Identify the formula to find the minimum sample size needed to achieve a specific standard deviation of the sample mean. The formula is $n = \left(\frac{\sigma}{\text{desired standard deviation}}\right)^2$ where $\sigma$ is the population standard deviation.
Plug in Values: Step $2$ : Plug in the values into the formula. Here, $\sigma = 6.0$ days (given as the standard deviation of the population), and the desired standard deviation of the sample mean is $0$ . $5$ day. So, $n = \left(\frac{6.0}{0.5}\right)^2$ .
Calculate Inside Parentheses: Step $3$ : Calculate the value inside the parentheses first. $\frac{6.0}{0.5} = 12.0$ .
Square Result: Step $4$ : Now, square the result from Step $3$ to find $n$ . $12.0^2 = 144$ .

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